Answer:
[tex]\sigma = 1.8[/tex]
Step-by-step explanation:
Given
[tex]\begin{array}{ccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} \ \\ P(x) & {0.2} & {0.1} & {0.1} & {0.2} & {0.2}& {0.2} \ \end{array}[/tex]
Required
The standard deviation
First, calculate the expected value E(x)
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
[tex]E(x) = 0 * 0.2 + 1 * 0.1 + 2 * 0.1 + 3 * 0.2 + 4 * 0.2 + 5 * 0.2[/tex]
[tex]E(x) = 2.7[/tex]
Next, calculate E(x^2)
[tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]E(x^2) = 0^2 * 0.2 + 1^2 * 0.1 + 2^2 * 0.1 + 3^2 * 0.2 + 4^2 * 0.2 + 5^2 * 0.2[/tex]
[tex]E(x^2) = 10.5[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{E(x^2) - (E(x))^2}[/tex]
[tex]\sigma = \sqrt{10.5 - 2.7^2}[/tex]
[tex]\sigma = \sqrt{10.5 - 7.29}[/tex]
[tex]\sigma = \sqrt{3.21}[/tex]
[tex]\sigma = 1.8[/tex] --- approximated
